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Question Number 206434 by MATHEMATICSAM last updated on 14/Apr/24

If tan^2 θ = 1 − x^2  then prove that  secθ + tan^3 θcosecθ = (√((2 − x^2 )^3 )) .

$$\mathrm{If}\:\mathrm{tan}^{\mathrm{2}} \theta\:=\:\mathrm{1}\:−\:{x}^{\mathrm{2}} \:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\mathrm{sec}\theta\:+\:\mathrm{tan}^{\mathrm{3}} \theta\mathrm{cosec}\theta\:=\:\sqrt{\left(\mathrm{2}\:−\:{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:. \\ $$

Answered by TonyCWX08 last updated on 14/Apr/24

secθ+tan^3 θ(((secθ)/(tanθ)))  =secθ+secθtan^2 θ  =secθ(1+1−x^2 )  =(√((1+tan^2 θ)))(2−x^2 )  =(√((2−x^2 )))(2−x^2 )  =(√((2−x^2 )^3 ))  Hence Proved.

$${sec}\theta+{tan}^{\mathrm{3}} \theta\left(\frac{{sec}\theta}{{tan}\theta}\right) \\ $$$$={sec}\theta+{sec}\theta{tan}^{\mathrm{2}} \theta \\ $$$$={sec}\theta\left(\mathrm{1}+\mathrm{1}−{x}^{\mathrm{2}} \right) \\ $$$$=\sqrt{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)}\left(\mathrm{2}−{x}^{\mathrm{2}} \right) \\ $$$$=\sqrt{\left(\mathrm{2}−{x}^{\mathrm{2}} \right)}\left(\mathrm{2}−{x}^{\mathrm{2}} \right) \\ $$$$=\sqrt{\left(\mathrm{2}−{x}^{\mathrm{2}} \right)^{\mathrm{3}} } \\ $$$${Hence}\:{Proved}. \\ $$

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